Question

From the circular disc of radius 4R two small disc of radius R are cut off.The center of mass of new structure will be:

• A. $i\frac{R}{5}+j\frac{R}{5}$
• B. $-i\frac{R}{5}+j\frac{R}{5}$
• C. $\frac{-3R}{14}(\hat{i}+\hat{j})$
• D. none of these

Answer 1

The correct option is C $\frac{-3R}{14}(\hat{i}+\hat{j})$

The new center of mass of a disc after releasing a small disc from it is given by:

$x_{cm}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$ where,

$m_1$ is the mass of original disc

$m_2$ is the mass of removed disc

$x_1$ is the center of mass position of original disc

$x_2$ is the center of mass position of removed disc

If we assume mass per unit area$=m$ then,

After removing the disk having c.o.m. at $(3R,0)$

mass of original disc $=m\times \pi \times (4R{)}^2=16m\pi R^2$

mass of removed disc$=m\pi R^2$

IN x-axis:

$x_{com}^{\prime }=\frac{16m\pi R^2\times 0-m\pi R^2\times \left(3R\right)}{16m\pi R^2-m\pi R^2}$

$=\frac{-3m\pi R^3}{15m\pi R^2}$

$=-\frac{1}{5}R$

IN y-axis:

$y_{com}^{\prime }=0-0=0$

Now,removing the disk with c.o.m at $(0,3R)$from the rest $15m\pi R^2$ mass disk:

$x_{com}^{\prime \prime }=\frac{15m\pi R^2\times x_{com}^{\prime }-m\pi R^2\times 0}{15m\pi R^2-m\pi R^2}$

$=\frac{-3R}{14}$

$y_{com}^{\prime \prime }=\frac{15m\pi R^2\times 0-m\pi R^2\times \left(3R\right)}{15m\pi R^2-m\pi R^2}$

$=\frac{-3m\pi R^3}{14m\pi R^2}$

$=-\frac{3}{14}R$

New center of position $=\frac{-3R}{14}(\hat{i}+\hat{j})$